# Prove Jensen's inequality

$$\left(\sum_{i=1}^na_i^p\right)^{1/p} \ge \left(\sum_{i=1}^na_i^q\right)^{1/q}$$ if $0 < p \le q$ for $a_i\ge 0$. I have proved that the inequality holds for $p=q$ (trivial) and I have also proved that it holds if one of the two sums is equal to 1, but I don't know how to continue. Please help.

• Also, it is trivial if the sums are $0$. In all other cases, how do they behave under scaling (multiplying all $a_i$ by some constant $c$)? – Daniel Fischer Nov 2 '14 at 21:07
• Have you heard about convexity? – Pedro Tamaroff Nov 2 '14 at 21:08
• no man i have not heard about it – Baquesh Nov 2 '14 at 21:09
• Prove Jensen's inequality - Why ? Has Jensen ever proved one of mine ? Course he hasn't ! And he expects me to prove one of his now, does he !? – Lucian Nov 2 '14 at 21:14

You want to prove that, for $0 < p \leq q <\infty$, $$\left(\sum_i \lvert{x_i}\rvert^q\right)^{1/q} = \lVert{x}\rVert_q \leq \lVert{x}\rVert_p = \left(\sum_i \lvert{x_i}\rvert^p\right)^{1/p}\;.$$ To see why, one can easily prove that if $\lVert{x}\rVert_p = 1$, then $\lVert{x}\rVert_q^q \leq 1$ (bounding each term $\lvert{x_i}\rvert^q \leq \lvert{x_i}\rvert^p$), and therefore $\lVert{x}\rVert_q \leq 1 = \lVert{x}\rVert_p$. Next, for the general case, apply this to $y = x/\lVert{x}\rVert_p$, which has unit $\lVert{\cdot}\rVert_p$ norm, and conclude by homogeneity of the norm.
• Replace my $x$'s by your $a$'s, and you obtain your question. What is confusing you? – Clement C. Nov 2 '14 at 21:33